How to assert function invocation order in jest

Question

I am mocking two functions with with jest.fn:

let first = jest.fn();
let second = jest.fn();

How can I assert that first called before second?

What I am looking for is something like sinon's .calledBefore assertion.

Update I used this simple "temporary" workaround

it( 'should run all provided function in order', () => {

  // we are using this as simple solution
  // and asked this question here https://stackoverflow.com/q/46066250/2637185

  let excutionOrders = [];
  let processingFn1  = jest.fn( () => excutionOrders.push( 1 ) );
  let processingFn2  = jest.fn( () => excutionOrders.push( 2 ) );
  let processingFn3  = jest.fn( () => excutionOrders.push( 3 ) );
  let processingFn4  = jest.fn( () => excutionOrders.push( 4 ) );
  let data           = [ 1, 2, 3 ];
  processor( data, [ processingFn1, processingFn2, processingFn3, processingFn4 ] );

  expect( excutionOrders ).toEqual( [1, 2, 3, 4] );
} );

Answer 1

Instead of your workaround you can install jest-community's jest-extended package which provides support for this via .toHaveBeenCalledBefore(), e.g.:

it('calls mock1 before mock2', () => {
  const mock1 = jest.fn();
  const mock2 = jest.fn();

  mock1();
  mock2();
  mock1();

  expect(mock1).toHaveBeenCalledBefore(mock2);
});

Note: per their doc you need at least v23 of Jest to use this function

https://github.com/jest-community/jest-extended#tohavebeencalledbefore

P.S. - This feature was added a few months after you posted your question, so hopefully this answer still helps!

Answer 2

The solution by clemenspeters (where he wanted to make sure logout is called before login) works for me:

const logoutSpy = jest.spyOn(client, 'logout');
const loginSpy = jest.spyOn(client, 'login');
// Run actual function to test
await client.refreshToken();
const logoutOrder = logoutSpy.mock.invocationCallOrder[0];
const loginOrder = loginSpy.mock.invocationCallOrder[0];
expect(logoutOrder).toBeLessThan(loginOrder)